Forest: I think your system (Bubble IRV, in the sense of bubble sort?) would have some good properties in terms of results. But honestly, I don't really see the point. We have a number of systems which give good results. To me, the point of designing new systems is to give good results while gaining some combination of (in roughly descending importance)<div>
<ul><li>Simplicity of explanation</li><li>Simplicity of counting (summability, etc.)</li><li>Simplicity of voting (ballot design, minimal strategy considerations, etc.)</li><li>Broad appeal (For instance, a method that would appeal to both IRV and Condorcet suppporters)</li>
<li>New, unexplored mathematical properties ("just for fun")</li></ul>I think your Bubble IRV doesn't really give any of the first three, and not too much of the last two.</div><div><br></div><div>It does, however, inspire me to present my own proposal, a simple modification of David's IRV3/AV3. My only change from his system is that equal ranking would be allowed. In the IRV part, equal-ranked-top would be counted as a full first-place vote for both (all) of the top candidates. </div>
<div><br></div><div>What does this do to summability (one of the main advantages of David's proposal over IRV)? It actually works fine. You'd keep three tallies: top ranks, approvals, and a Condorcet matrix. Since with three candidates there's only one true IRV elimination, equal-ranking doesn't cause the logistical headaches it would with full IRV.</div>
<div><br></div><div>What does it do to simplicity of explanation? In my opinion, it's at least as good as its predecessor IRV3/AV3. "Keep the three top approvals; discard the one with fewest top ranks; and use preferences to see who'd win between the remaining two." In a certain sense, that's actually simpler than even giving a full explanation of IRV.</div>
<div><br></div><div>What does it do to simplicity of voting? It is much better for ballot design. With equal-ranking allowed, you'd simply eliminate the problem of spoiled ballots, which seems to me to be a real concern for IRV. And allowing equality makes it possible to vote a ratings-style ballot, which is cognitively easier.</div>
<div><br></div><div>How about strategy? Additional votes at level 2 or 3 would essentially be a way to make the approval part of this method freer, without the arbitrary limit of three approvals. Additional votes at top level would be a way for a solid majority coalition to ensure that their principal enemy is eliminated. The chicken dilemma still applies, so it would rarely (never?) be "strategically optimal" from a first-order perspective to vote two candidates at top ranking; but on the whole, I think it's good to allow voters the option to explicitly say that they don't care about such first-order strategic considerations between two candidates they consider to be clones.</div>
<div><br></div><div>In terms of results, I think this system would tend to give the IRV winner, with perhaps a small step in the direction of MJ. That's not my favorite place to be; it still allows for self-perpetuating two-party domination. However, I think that there's a real possibility that this system would allow for smoother transitions if the set of top two parties changed at a local level. So ... well, perhaps it's my "I invented it" bias, but I think it's good enough.</div>
<div><br></div><div>Thoughts?</div><div><br></div><div>Jameson</div><div><br></div><div>ps. One more minor comment on Forest's Bubble IRV proposal, below:</div><div><br><div class="gmail_quote">2011/11/5 <span dir="ltr"><<a href="mailto:fsimmons@pcc.edu" target="_blank">fsimmons@pcc.edu</a>></span><br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Dear EM Folks,<br>
I’ve been very busy, while watching postings on the EM list out of the corner of my eye. I was very<br>
happy to notice Mike Ossipoff’s interesting contributions. In particular his promotion of a variant of IRV<br>
where equal rank counts whole strikes me as promising in the context of the “chicken problem.”<br>
<br>
On a related note I have been thinking about how to make a monotone variant of IRV. Perhaps the two<br>
ideas could be combined without sacrificing all of the other nice features that IRV(=whole) seems to have.<br>
<br>
It is well known that certain kinds of elimination methods cannot satisfy the monotonicity criterion. The<br>
basic variant of IRV is of that type, namely what I call the “restart with each step type” of elimination<br>
method. This means that when one candidate is eliminated, the next stage starts all over again without<br>
learning from or memory of the eliminated candidate. Range elimination methods that renormalize all of<br>
the ballots at each stage are of this type, too, since the renormalization is an attempt to eliminate the<br>
effect of the eliminated candidates on the remaining stages of the process.<br>
<br>
But some methods of elimination that do not suffer from the “restart problem” turn out to be monotone.<br>
For example, approval elimination where the original approvals are kept throughout the whole elimination<br>
process; trivially the highest approval candidate is the last one left.<br>
<br>
Now here’s what I propose for an IRV variant:<br>
<br>
1. Use the ranked ballots to find the pairwise win/loss/tie matrix M. This matrix stays the same<br>
throughout the process.<br>
<br>
2.Initialize a variable U (for Underdog) with the name of the candidate ranked first on the fewest number<br>
of ballots, and eliminate U from the ballots.<br>
<br>
3.While more than one candidate remains, eliminate candidate X that is ranked first on the fewest<br>
number of ballots after the previously eliminated candidates’ names have been wiped from the ballots (as<br>
in IRV elimination) and then replace U with X, unless U defeats X, in which case leave the value of U<br>
unchanged.<br>
<br>
4. Elect the pairwise winner between the last value of U and the remaining candidate.<br>
<br>
Note that a simplified version of this where you just eliminate the pairwise loser of the two candidates<br>
ranked first on the fewest number of ballots in NOT monotone. We have to remember the previous<br>
survivor and carry him/her along as "underdog challenger" to make this method monotone.<br>
<br>
Note also that this method satisfies the Condorcet Criterion. So we gain monotonicity and CC, but what<br>
desireable criteria do we lose? It still works great on the scenario<br>
<br>
49 C<br>
27 A>B<br>
24 B<br>
<br>
Candidate A starts out as underdog, survives B, and is beaten by C, so C wins. </blockquote><div><br></div><div>Wouldn't B be the underdog initially here? (Not that it matters to the result or to the further analysis below.)</div>
<div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">But if B supporters<br>
really prefer A to C they can make A win. On the other hand if the A supporters believe that the B<br>
supporters are indifferent between A abd C, they can vote A=B, so that B wins.<br>
<br>
When I have more time, I'll sketch a proof of the monotonicity.<br>
<br>
Comments?<br>
<br>
Thanks,<br>
<br>
Forest<br>
----<br>
Election-Methods mailing list - see <a href="http://electorama.com/em" target="_blank">http://electorama.com/em</a> for list info<br>
</blockquote></div><br>
</div>